In second quantization, the Hamiltonian describing the hopping process between two neighboring sites is given ($N$ - number of particles and $M$ - number of sites) by:

$$\hat{\mathcal H} = J\sum\limits_{\langle i,j \rangle}\hat{a}^{\dagger}_{i}\hat{a}_{j}$$

It can be diagonalized using Fourier series

$$\hat{a}_{i} = \frac{1}{\sqrt{M}}\sum\limits_{\mathbf{k}} \hat{b}_{\mathbf{k}}e^{-i\mathbf{k}\cdot\mathbf{R}_i}$$

The ground state is given by

$$|\mathrm{GS}\rangle = \frac{1}{\sqrt{N!}}\left( \hat{b}^{\dagger}_{\mathbf{k} = \mathbf{0}}\right)^N | 0 \rangle \approx C\ e^{\sqrt{N}\hat{b}^{\dagger}_{\mathbf{k} = \mathbf{0}}}| 0 \rangle$$

How is the approximation justified ($C$ ensures normalization)?


I calculated a few quantities using both states. First of all both states can be represented using site creation operators $\hat{a}_i$ in the following way: $$|\rm GS_{1}\rangle = \frac{1}{\sqrt{N!}}\left[\frac{1}{\sqrt{M}}\sum\limits_{i=1}^{M}\hat{a}_{i}^{\dagger} \right]^{N} |0\rangle$$ $$|\rm GS_2 \rangle = \prod\limits_{i=1}^{M}e^{\sqrt{\frac{N}{M}}(\hat{a}^{\dagger}_i - \hat{a}_i)}|0\rangle$$ so the first state is $\rm SU(M)$ coherent state while the second one is the product of Glauber coherent states. A few quantities of interest:

  1. Energy - expectation value of the Hamiltonian: $$\langle \hat{\mathcal H} \rangle_{1} = 2JN$$ $$\langle \hat{\mathcal H} \rangle_{2} = 2JN$$

  2. Particle density and fluctuations: $$\langle \hat{n}_{i} \rangle_{1} = \frac{N}{M},\ \ \ \langle \Delta \hat{n}_{i}^{2} \rangle_{1} = \frac{N}{M}\left(1 - \frac{1}{M} \right)$$ $$\langle \hat{n}_{i} \rangle_{2} = \frac{N}{M},\ \ \ \langle \Delta \hat{n}_{i}^{2} \rangle_{2} = \frac{N}{M}$$

  3. Total number of particles and fluctuations: $$\langle \hat{N} \rangle_{1} = N,\ \ \ \langle \Delta \hat{N}^{2} \rangle_{1} = 0$$ $$\langle \hat{N} \rangle_{2} = N,\ \ \ \langle \Delta \hat{N}^{2} \rangle_{2} = N$$

As @NorberSchuch mentioned in his answer below particle density fluctuations are practically the same and coincide in the thermodynamic limit.


2 Answers 2


The basic idea is that the term with the highest weight in the exponential series is exactly the desired term. Further, all of the weight is in terms with closeby particle number, and small fluctuations in this number do not matter in many cases.

(Note: I will omit the $k=0$ subscript and just write $b^\dagger$ in the following for convenience.)

First, we have that
$$ e^{\sqrt{N}b^\dagger} |0\rangle = \sum \frac{\sqrt{N}^k}{k!} (b^\dagger)^k |0\rangle\ . $$ Let us now study which term in the series has the highest weight. To this end, we have to make sure that the corresponding states have the same normalization. This is, if we define a normalized state $$ |k\rangle = \frac{(b^\dagger)^k}{\sqrt{k!}}|0\rangle\ , $$ each of which appears in the series with a probability $$ |c_k|^2 = \frac{N^k}{k!}\ . $$ This is an (unnormalized) Poisson distribution, whose mean is exactly at $k=N$, as desired.

Note that this does in fact not mean that the two states are close in any distance measure (in fact, they aren't). However, it tells us that most of the weight in the sum is in states with $|k-N|\le c\sqrt{N}$, as the variance of the Poisson distribution is $N$. Since in many applications, we in fact care about the number of particles per site (which is an intensive quantity and well-defined as the system size goes to infinity), and any such fluctuation in $k$ will vanish as $N\rightarrow\infty$, both states will have the same particle density in the thermodynamic limit, and thus describe the same physics.

  • $\begingroup$ I performed calculations with both states and they have the same energy and mean number of particles per site. So, I don't understand your answer cause you wrote that in the thermodynamic limit the particle density is the same. $\endgroup$
    – WoofDoggy
    Apr 13, 2016 at 8:23
  • $\begingroup$ @Nex_Friedrich "particle density" = "mean number of particles per site". It sounds to me like we are saying the same. Where do you see a problem? $\endgroup$ Apr 13, 2016 at 8:55
  • $\begingroup$ your answer: "both states will have the same particle density in the thermodynamic limit, and thus describe the same physics" - they have the same density for any $N$. $\endgroup$
    – WoofDoggy
    Apr 13, 2016 at 10:36
  • $\begingroup$ both states are symmetric under exchange of creation operators $\hat{a}^{\dagger}_{i}$, so they will have the same mean number of particles and fluctuations per site. $\endgroup$
    – WoofDoggy
    Apr 13, 2016 at 10:40
  • $\begingroup$ You are absolutely right (and indeed, the mean of the distribution is the right one for any system size). The point I wanted to make was indeed that the fluctiations of the density become small in the thermodynamic limit. But it might well be that even stronger statements hold -- feel free to edit my answer accordingly! $\endgroup$ Apr 13, 2016 at 11:35

1) Let's talk about the energy of ground states. With notations $$ |\text{GS}_{1}\rangle \equiv \frac{1}{\sqrt{N!}}\left(\hat{b}^{\dagger}_{\mathbf k = 0}\right)^{N}|0\rangle , \quad |\text{GS}_{2}\rangle \equiv Ce^{\sqrt{N}\hat{b}_{\mathbf k = 0}^{\dagger}}|0\rangle $$ you obtain $$ \hat{H}|\text{GS}_{1}\rangle =J \hat{b}^{\dagger}_{\mathbf k = 0}\hat{b}_{\mathbf k =0 }\left(\hat{b}^{\dagger}_{\mathbf k = 0}\right)^{N}|0\rangle = JN|\text{GS}_{1}\rangle, $$ $$ \hat{H}|\text{GS}_{2}\rangle =J \hat{b}^{\dagger}_{\mathbf k = 0}\hat{b}_{\mathbf k =0 }|\text{GS}_{2}\rangle = \left| \hat{b}^{\dagger}_{\mathbf k = 0}|\text{GS}_{2}\rangle = \hat{b}^{\dagger}_{\mathbf k = 0}|\text{GS}_{2}\rangle = \sqrt{N}|\text{GS}_{2}\rangle\right| = JN|\text{GS}_{2}\rangle $$ That's why you may talk about similarity of $|\text{GS}_{1}\rangle$ and $|\text{GS}_{2}\rangle$: they have the same energy. Also they have the same number density $n \equiv \frac{N}{M}$.

2) However, they are not equivalent at least since for the planar wave $$ \hat{\psi}_{i} = \frac{1}{\sqrt{M}}\sum_{\mathbf k}e^{i\mathbf k \cdot \mathbf R_{i}}\hat{b}_{\mathbf k} $$ corresponding VEVs are $$ \psi^{(1)} \equiv \langle \text{GR}_{1}|\hat{\psi}_{i}|\text{GR}_{1}\rangle = 0, \quad \psi^{(2)}\equiv \langle \text{GR}_{1}|\hat{\psi}_{i}|\text{GR}_{1}\rangle = \sqrt{\frac{N}{M}} \neq 0 $$


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